Valve Sizing Examples

Application Notes

Valve sizing sample problems

General informationThese application notes provide a simplified procedure forsizing control valves at standard service conditions.The data required to size a valve, such as the nominal size,nominal pressures and KVS, is contained in the SAMSON datasheets for self-operated regulators and control valves.Self-operated regulators and control valves can be sizedaccurately using the DIN IEC 534 procedure. For mostapplications, however, the following sizing equationsformulated in the VDI/VDE Guideline 2173 provide a sufficientdegree of accuracy.In order to calculate the valve flow coefficient KV, the operatingdata specified in the figure to the right is required.

Typical sizing coefficientsExplanatory notesControl valves and self-operated regulatorsRated travel · The rated travel H100 is the amount of movementof the valve closure member from the closed position to the des-ignated full open position published by the manufacturer foreach control valve series.

KV · The valve flow coefficient KV is defined as the number ofcubic meters per hour (volume flow rate V

.) of 5 to 30 °C water

that will flow through a control valve at a specified travel H with adifferential pressure (Δp = p1 – p2) of 105 Pa (1 bar) across it.

KVS · The KVS value is the expected flow coefficient KV of thevalve at rated travel H100 indicated and published by the manu-facturer for each valve type (series).

KV100 · The KV100 value is the effective (actual) flow coefficientKV of the valve at rated travel H100. It must not deviate by morethan ±10 % from the indicated KVS value.Self-operated regulatorsSelf-operated flow regulators can only be sized if the upstreampressure p1 is constant.

Safety factor S =KK

VS

V

For self-operated regulators: S ≈ 1.3 to 5

To ensure the proper operation of a self-operated regulator, thekinematic viscosity ν if the medium to be controlled must notexceed 1 · 10-4 m2/s = 100 cSt.

Conversion of the maximum volume flow rate for liquidswith different densities

V.

B = V.

A ⋅ ρρ

A

B

⎛

⎝⎜

⎞

⎠⎟

Edition May 2005

Application Notes AB 04 EN

Fig. 1 · Operating data for determining the KV value

p1 p2

Q W

H100H0

t1

p1 Upstream pressure in barp2 Downstream pressure in barΔp Differential pressure (p1 – p2) across the valve in barH Travel in mmV.

Volume flow rate in m3/hW Mass flow rate in kg/hρ Density (general) in kg/m3

ρ1 Upstream density (for gases and steam) in kg/m3

t1 Upstream temperature in ° C

Contents PageValve sizing for liquid service 2Valve sizing for steam service 5Valve sizing for gas service 7Valve sizing for air service 9Flow through pipelines 11

ρ Density in kg/m3

V.

Volume flow rate in m3/h

KVS: Indicated KVSKV: Calculated KV

Equation (1) generally applies to liquids:

Diagram 1 shows the relationship between V., KV and Δp for liquids that have

a density of ρ = 1000 kg/m3 at a temperature of t = 20 °C.

Flow regulators · Differential pressure across the valveCalculate the minimum required differential pressure across the valve usingEquation (2):

Determine: Min. differential pressure Δp across fully open valveGiven: Type 42-36 Self-operated Flow Regulator, diff. pressure

across restriction 0.2 bar, DN 40, KVS 20, volume flow rateSolution: Calculate Δpmin using Equation (2):

Δprestriction = 0.2 to 0.5 bar depending on regulator version

Determine: Volume flow rate of acetone (m3/h) with valve fullyopen

Given: Type 3241-1 Pneumatic Control Valve, DN 40,differential pressure Δp = p1 – p2 · Density of acetonein kg/m3

Solution:Calculate V

.using

Equation (1):

Determine: Type … Pressure Reducing Valve, KVS , nominal valve sizeGiven: Volume flow rate of water · Differential pressure · Density

of water ρ in kg/m3

Solution: Calculate the KV value using Equation (3) derived fromEquation (1):

Determine the KVS value from the following equation usingthe calculated KV value. In general: KVS ≈ 1.3 · KVSolution provided by Diagram 1:

For Δp = 2.1 bar and V.

= 12 m3/h, a KV value of ap-proximately 8.2 m3/h is indicated in Diagram 1.The value for the flow velocity in the pipe can be found inDiagram 4 on page 11 using V

.= 12 m3/h and DN 40:

wpipe ≈ 2.8 m/s

22 AB 04 EN

V.

= KV ·1000 ⋅ Δp

ρ(1)

This equation containsthe dimensional factor1.

Symbols and unitsp1 Absolute pressurep2 in barΔp In barρ Density in kg/m3

V.

Volume flow rate in m3/hKV KV in m3/hΔpmin Minimum differential pressure across the

valve in barΔprestriction Differential pressure in bar created

deliberately across the restriction in theflow path for flow measurement purposes

KVS Indicated valve sizing coefficient in m3/h

}

Liquid service

Sizing sample problem 2

V.

= KVS ·1000 ⋅ Δp

ρ(1)

KVS = 25 m3/h 1)

Δp = p1 – p2 = 0.5 barρ = 800 kg/m3

V.

= 25 · 1000 0 5800

⋅ . = 19.76 m3/h

1) KVS and hence also V., the calculated volume flow

rate, have a permissible tolerance of ±10 %.


KV = V.

·ρ

1000 ⋅ Δp(3)

V.

= 12 m3/hΔp = p1 - p2 = 2.1 barρ = 1000 kg/m3

KV = 12 · 10001000 21⋅

=.

8.2 m3/h

KVS = 1.3 · KV = 1.3 · 8.2 = 10.7 m3/h

Selection: Type 41-23, DN 40, KVS = 20

Determining S, the safety factor:

S = KK

VS

V= 20

82.≈ 2.4

Δpmin = Δprestriction +V

K

.

VS

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

(2)


KVS = 20 m3/hΔrestriction = 0.2 bar

V.

= 10 m3/h

Δpmin = 0.2 + 1020

⎛⎝⎜

⎞⎠⎟

2= 0.45 bar

Δmin = Δprestriction +V

K

.

VS

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

(2)

AB 04 EN 3

Diagram 1 · Volume flow rate diagram for water with ρ = 1000 kg/m3, t = 20 °C

V.

in m3/hKV in m3/h

Δp in bar V.

= KV · 1000 ⋅ Δpρ

10 20 30 40 60 100800.01

0.02

0.03

0.040.050.06

0.080.1

0.4

0.2

0.3

10.8

0.60.5

4

2

3

108

65

40

20

30

10080

6050

400

200

300

1000

Kvm3

h

800

600500

200 300 400 1000600 800 mbar

0.1 0.2 0.60.40.3 10.8 1042 3 6 8 20 30 40 bar

p

0.0100.0120.0160.0200.0250.0300.040.050.060.080.100.12

0.160.200.250.300.4

0.50.6

0.81.0

1.21.6

2.02.5

3.04

56

810

1216

2025

3040

5060

80100120

160200

250300400

5006008001000

Determine: Differential pressure Δp = p1 – p2 with valve fully openGiven: Type 4 Self-operated Temperature Regulator with DN 50,

V., volume flow rate of water,

ρ, density of water in kg/m3

Solution: Determine the differential pressure using Equation (4)derived from Equation (1):

Solution provided by Diagram 1:

For V.

= 10 m3/h and KVS 32, a differential pressure ofΔp ≈ 0.1 bar is obtained from Diagram 1.

44 AB 04 EN

Liquid service


Δp =V

K

.

VS

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2·

ρ1000

(4)

KVS = 32 m3/h

V.

= 10 m3/hρ = 1000 kg/m3

Δp = 1032

⎛⎝⎜

⎞⎠⎟

2· 1000

1000= 0.1 bar

A modified version of Equation (8) applies to steam:

Z The dimensionless compressibility factor Z is definedas follows: Z = 14.2 · p1 ⋅ρ1. Determine Z from Table 2using the upstream pressure p1 and differentiating be-tween saturated steam and superheated steam.

m Determine the dimensionless head loss coefficient m fromTable 1 or, for in-between values, from Diagram 2 using( = 1.135.

Determine: Steam mass flow rate in kg/h with valve fully openGiven: Type 3241-2 Electric Control Valve · Steam

temperature · Upstream and downstream pressures

Solution: Calculatepp

2

1(convert into absolute pressure if required)

Determine m from Table 1 or Diagram 2Determine Z from Table 2 using the upstream pressure andtemperature

Determine: Type … Pressure Reducing Valve, KVS value, nominal sizeGiven: Steam mass flow rate · Steam temperature · Upstream

and downstream pressures


2


Determine m from Table 1 · Determine Z from Table 2using the upstream pressure and temperatureCalculate KV from Equation (6):

Determine the KVS value of the valve from the followingequation using the calculated KV value.In general: KVS ≈ 1.3 · KV

Determine: Differential pressure Δp = p1 – p2 across fully open valveGiven: Type 4 Self-operated Temperature Regulator · Steam mass

flow rate · Steam temperature · Upstream pressureSolution: Determine Z from Table 2 using the upstream pressure and

temperature

Using m = 0.701, determine the ratio ofpp

2

1from Diagram 2

p2 = 0.884 · p1 ⇒ Differential pressure Δp = p1 – p2

AB 04 EN 5

Steam service

W = KV · m · Z (5)

Symbols and unitsp1 Absolute pressurep2 in barΔp Differential pressure in barW Mass flow rate in kg/hKV KV value in m3/hm Head loss coefficient, dimensionlessZ Compressibility factor, dimensionless

}


W = KVS · m · Z (5)

KVS = 35 m3/h 1)

t = 200 °Cp1 = 4 bar p2 = 3 barpp

2

1= 3

4= 0.75

m = 0.92Z = 38.5W = 35 · 0.92 · 38.5 = 1240 kg/h

1) KVS and hence also W, the calculated mass flowrate, have a tolerance of ±10 %.


KV =W

Z m⋅(6)

W = 1000 kg/ht = corresponds to saturated steamp1 = 7 bar p2 = 2 barpp

2

1

= 27

= 0.286

m = 1Z = 71.3

KV = 1000713 1. ⋅

= 14 m3/h

KVS = 1.3 · KV = 1.3 · 14 = 17.5 m3/h

Selection: Type 39-2, DN 40, KVS = 20


m =W

Z KVS⋅(7)

KVS = 20 m3/hW = 1000 kg/ht = corresponds to saturated steamp1 = 7 barZ = 71.3

m = 1000713 20. ⋅

= 0.701

pp

2

1= 0.884

p2 = 0.884 · 7 = 6.2 barΔp = p1 – p2 = 7 – 6.2 = 0.8 bar

6 AB 04 EN6 AB 04 EN

Table 1 · Head loss coefficient m as a function of p2/p1

Pressure ratio p2/p1 0 to 0.6 0.70 0.75 0.80 0.85 0.90 0.95 0.99

Head loss coefficient m 1.0 0.96 0.92 0.86 0.77 0.66 0.48 0.22

Table 2 · Compressibility factor Z for steam · All pressures are indicated as absolute pressures in bar

Compressibility factor Z for ...

p1in bar

Saturatedsteam

Superheated steam at the following temperatures ...60 °C 80 °C 100 °C 120 °C 140 °C 160 °C 180 °C 200 °C 250 °C 300 °C 350 °C 400 °C

0.1 1.16 1.13 1.1 1.07 1.04 1.02 0.99 0.97 0.95 0.90 0.86 0.83 0.800.2 2.27 2.27 2.21 2.15 2.09 2.04 1.99 1.95 1.90 1.81 1.73 1.66 1.590.3 3.37 3.31 3.22 3.14 3.06 2.99 2.92 2.86 2.71 2.59 2.49 2.390.4 4.45 4.42 4.29 4.18 4.08 3.98 3.89 3.81 3.62 3.46 3.32 3.190.5 5.53 5.37 5.23 5.10 4.98 4.86 4.76 4.52 4.33 4.15 3.990.6 6.58 6.45 6.28 6.12 5.97 5.84 5.72 5.43 5.19 4.98 4.780.7 7.65 7.53 7.33 7.15 6.97 6.82 6.67 6.34 6.06 5.80 5.590.8 8.71 8.62 8.39 8.17 7.97 7.79 7.63 7.25 6.91 6.64 6.370.9 9.76 9.70 9.44 9.19 8.98 8.77 8.58 8.16 7.90 7.37 7.181.0 10.8 10.8 10.5 10.2 9.98 9.76 9.53 9.07 8.66 8.30 7.981.1 11.9 11.5 11.3 11.0 10.8 10.5 10.0 9.50 9.10 8.701.2 12.9 12.6 12.3 12.0 11.8 11.4 10.9 10.4 10.0 9.601.3 13.9 13.7 13.3 13.0 12.7 12.3 11.8 11.2 10.8 10.41.4 15.0 14.7 14.3 14.0 13.7 13.4 12.7 12.1 11.6 11.21.5 16.0 15.8 15.4 15.0 14.7 14.3 13.6 13.0 12.4 12.01.6 17.0 16.9 16.4 16.0 15.6 15.3 14.5 13.9 13.3 12.81.7 18.0 17.9 17.5 17.0 16.6 16.3 15.4 14.7 14.1 13.61.8 19.1 19.0 18.5 18.0 17.6 17.2 16.4 15.6 14.9 14.41.9 20.1 20.1 19.5 19.0 18.6 18.1 17.3 16.5 15.8 15.22.0 21.1 21.1 20.6 20.0 19.6 19.1 18.2 17.3 16.6 16.12.2 23.2 22.6 22.1 21.5 21.0 20.0 19.1 18.3 17.62.4 25.2 24.7 24.1 23.5 23.1 21.8 20.8 20.0 19.22.6 27.2 26.8 26.0 25.5 24.9 23.6 22.6 21.5 20.82.8 29.3 28.9 28.1 27.5 26.8 25.5 24.3 23.2 22.43.0 31.0 31.0 30.2 29.4 28.8 27.3 26.0 24.9 24.03.2 33.4 33.1 32.2 31.4 30.7 29.1 27.8 26.6 25.63.4 35.4 35.2 34.3 33.4 32.6 31.0 29.6 28.2 27.23.6 37.4 37.3 36.3 35.4 34.6 32.8 31.3 29.9 28.93.8 39.4 38.3 37.4 36.5 34.7 33.0 31.6 30.44.0 41.4 40.4 39.4 38.5 36.5 35.1 33.3 32.04.5 46.4 45.6 44.4 42.8 41.1 39.1 37.3 36.15.0 51.4 50.8 49.4 48.2 45.7 43.6 41.8 40.05.5 56.4 56.0 54.4 53.0 50.2 47.8 46.7 44.26.0 61.4 61.2 59.5 57.9 54.9 52.3 50.2 48.26.5 66.3 64.6 62.9 59.4 56.6 54.2 52.27.0 71.3 69.7 67.8 64.2 61.1 58.3 56.28.0 81.2 79.9 77.6 73.4 69.8 67.0 64.39.0 91.0 90.2 87.7 82.6 78.7 75.0 72.410.0 101 101 97.9 92.2 87.4 83.2 80.411.0 111 108 102 96.5 92.1 88.512.0 121 118 111 105 99.7 96.713.0 130 128 121 114 109 10514.0 140 139 130 123 118 11315.0 150 150 139 132 125 12116.0 160 149 141 134 12917.0 170 159 150 143 13718.0 180 169 159 151 14619.0 189 178 168 161 15420.0 199 188 177 168 16221.0 209 198 187 178 17023.0 229 218 205 195 18725.0 248 238 224 213 20327.0 268 258 242 230 21629.0 288 279 261 248 23631.0 308 300 280 264 25333.0 328 322 299 282 27035.0 348 343 318 301 28637.0 368 365 338 319 30439.0 388 387 356 337 32041.0 408 376 354 338

Determine: Propane gas flow rate W in kg/h with valve fully openGiven: Type 3241-1 Pneumatic Control Valve, DN 50 ·

Upstream and downstream pressuresSolution: Determine the upstream density ρ1 from Diagram 3

Calculatepp

2

1· Using

pp

2

1and ( = 1.3, determine m from

Diagram 2.

Determine: Type … Pressure Reducing Valve, KVS value, nominal sizeGiven: Nitrogen system · Nitrogen flow rate · Pressures p1 and p2


2

1· Determine m from Diagram 2

(( =1.4) · Determineρ1 from Diagram 3 using p1 = 5 bar

Determine the KVS value from the following equation usingthe calculated KV value. In general: KVS ≈ 1.3 · KV

Determine: Differential pressure Δp = p1 – p2 across fully open valveGiven: Type 241-2 Electric Control Valve, DN 20 · Nitrogen

system · Nitrogen flow rate · Upstream pressureSolution: Determine ρ1 from Diagram 3 using p1 = 5 bar

Use m to determinepp

2

1

from Diagram 2

AB 04 EN 7

Diagram 2 · Head loss coefficient m as a function of pp

2

1

0.60.6

0.2

0.7

0.3 0.4 0.5

0.8

0.9

1

p2p1

0.80.7 0.9 1 m

( = 1.66 For monatomic gases such ashelium, argon, krypton

( = 1.4 For diatomic gases such ashydrogen, nitrogen, air,chlorine gas, town gas

( =1.3 For triatomic and multiatomic gasessuch as carbon dioxide, propane,butane, methane, acetylene, ammonia

( = 1.135 For steam

( = 1.135

( = 1.3

( = 1.4

( = 1.66

Gas service

Symbols and unitsp1 Absolute pressurep2 in barΔp Differential pressure in barW Mass flow rate in kg/hKV KV value in m3/hρ1 Density in kg/m3

The calculation procedure provides approximatevalues meeting practical needs.

The values for m and ρ1can be found in Diagrams 2and 3 respectively.

}

W = 14.2 · KV · m · p1 1⋅ρ (8)


KVS = 35 m3/h 1)

p1 = 2.7 bar p2 = 2.2 barρ1 = 5.28 kg/m3

pp

2

1= 22

27..

= 0.815

m = 0.805W = 14.2 · 35 · 0.805 · 27 528. .⋅ = 1511 kg/h


W = 14.2 · KVS · m · p1 1⋅ρ (8)


KV =W

14.2 m p1 1⋅ ⋅ ⋅ρ(9)

p1 = 5 bar p2 = 3 barW = 250 kg/hpp

2

1

= 35

= 0.6

m = 0.99ρ1 = 6.2 kg/m3

KV = 250142 099 5 62. . .⋅ ⋅ ⋅

= 3.19 m3/h

KVS = 1.3 · KV = 1.3 · 3.19 = 3.99 m3/h

Selection: Type 44-1 Pressure Regulator, G ¾, KVS = 4


m =W

14.2 K pVS 1 1⋅ ⋅ ⋅ρ(10)

p2 = 0.83 · p1Δp = p1 – p2

KVS = 4 m3/hp1 = 5 barW = 250 kg/hρ1 = 6.2 kg/m3

m = 250142 4 5 62. .⋅ ⋅ ⋅

= 0.791

pp

2

1

= 0.83

p2 = 0.83 · 5 = 4.15 barΔp = 5 – 4.15 = 0.85 bar

Forpp

2

1< 0.6 ⇒m = 1

8 AB 04 EN

Diagram 3 · Density of gases ρ or ρ1 as a function of pressure at 0 °C

If the operating temperature t deviates considerably from 0 °C,correct ρ1 using the equation ρt = ρ1 ⋅

273273 t+

1 Chlorine gas 5 Air 9 Methane2 Butane 6 Nitrogen 10 Town gas3 Propane 7 Acetylene 11 Hydrogen4 Carbon dioxide 8 Ammonia

0.01 0.02 0.03 0.06 0.10.04 0.08

0.001

0.002

0.003

0.0040.0050.006

0.0080.01

0.04

0.02

0.03

0.10.08

0.060.05

0.4

0.2

0.3

10.8

0.60.5

4

2

3

108

65

40

20

30

100

kg

m3

80

6050

0.2 0.3 0.4 10.6 0.8 2 3 4 20 30 40106 8

ρ

1p

bar

1

2

3

8

456

7910

11

The calculation procedure provides approximate values meeting practicalneeds. The equation derived for dry air is as follows:

Equation (10) can be written as follows:

The value for Z can be obtained from Table 4 using p1. Thevalue for m can be found in Table 3 (in-between valuescan be determined from Diagram 2 using ( = 1.4)

Determine: Mass flow rate in kg/h with valve fully openGiven: Type 41-23 Self-operated Pressure Regulator,

DN 50 · Upstream and downstream pressures · Temperature


2


Determine the value for m from Table 3 or Diagram 2Determine the value for Z from Table 4 using the upstreampressure and temperature

Determine: Type … Pressure Reducing Valve, KVS value, nominal sizeGiven: Upstream and downstream pressures · Temperature ·

Process medium is compressed air


2


Determine the value for m from Table 3 or Diagram 2 ·Determine the value for Z from Table 4 using the upstreampressure and temperature

Determine the KVS value from the following equation usingthe calculated KV value. In general: KVS ≈ 1.3 · KV

Determine: Differential pressure Δp = p1 – p2 across fully open valveGiven: Type 42-24 Differential Pressure Regulator, DN 50 · Flow

rate of compressed air · Upstream pressure · TemperatureSolution: Determine the value for Z from Table 4 using the upstream

pressure and temperature

Determinepp

2

1for m = 0.884 from Diagram 2 or Table 3

p2 = 0.75 · p1 ⇒ Differential pressure Δp = p1 – p2

AB 04 EN 9

Air service

W = 15.3 · m · KV · p1 1⋅ρ (11)

Symbols and unitsp1 Absolute pressurep2 in barΔp Differential pressure in barW Mass flow rate in kg/hKV KV value in m3/hρ Density in kg/m3

m Head loss coefficient, dimensionlessZ Compressibility factor, dimensionless

Note:Please contact SAMSON AG in Frankfurt/Main to obtainmore detailed information on sizing flow regulators forair service.

}


W = KV · m · Z (12)

W = KVS · m · Z (12)

KVS = 32 m3/h 1)

t = 20 °Cp1 = 4 bar p2 = 3 barpp

2

1= 3

4= 0.75

m = 0.884Z = 66W = 32 · 0.884 · 66 = 1867 kg/h



KV =W

Z m⋅(13)

p1 = 5 bar p2 = 3 bart = 20 °CW = 190 kg/hpp

2

1

= 35

= 0.6

m = 0.982Z = 82.60

KV = 1900982 82 60. .⋅

= 2.34 m3/h

KVS = 1.3 · KV = 1.3 · 2.34 = 2.93 m3/h

Selection: Type 44-1, G ½, KVS = 3.2


m =W

Z KVS⋅(14)

KVS = 32 m3/hW = 3270 kg/ht = 20 °Cp1 = 7 barZ = 115.6

m = 3270115 6 32. ⋅

= 0.884

pp

2

1= 0.75

p2 = 0.75 · 7 = 5.25 barΔp = p1 – p2 = 7 – 5.25 = 1.75 bar

10 AB 04 EN

Table 3 · Head loss coefficient m as a function of p2/p1 · All indicated pressures are absolute pressures in bar

Pressure ratio p2/p1 0.527 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 0.99

Head loss coefficient m 1 0.982 0.978 0.935 0.884 0.818 0.730 0.623 0.448 0.207

Table 4 · Compressibility factor Z for air · All pressures are indicated as absolute pressures in bar

Compressibility factor Z for ...

p1in bar

Dry air at the following temperatures ...0 °C 20 °C 50 °C 100 °C 150 °C 200 °C 250 °C 300 °C

0.1 1.71 1.65 1.57 1.47 1.38 1.30 1.24 1.180.2 3.42 3.30 3.15 2.93 2.77 2.60 2.47 2.310.3 5.13 4.96 4.74 4.39 4.13 3.89 3.71 3.550.4 6.84 6.61 6.29 5.85 5.50 5.20 4.95 4.720.5 8.55 8.26 7.87 7.32 6.88 6.50 6.18 5.920.6 10.26 9.90 9.42 8.79 8.24 7.79 7.42 7.090.7 11.97 11.56 11.00 10.22 9.61 9.09 8.64 8.270.8 13.68 13.22 12.58 11.72 11.00 10.40 9.79 9.450.9 15.40 14.86 14.15 13.18 12.36 11.69 11.12 10.621.0 17.10 16.50 15.72 14.65 13.75 13.00 12.35 11.811.1 18.83 18.15 17.30 16.07 15.10 14.32 13.60 13.001.2 20.50 19.80 18.20 17.52 16.50 15.60 14.70 14.121.3 22.10 21.42 20.45 18.25 17.85 16.90 16.09 15.351.4 24.00 23.10 22.00 10.70 19.25 18.19 17.30 16.551.5 25.65 24.75 23.60 21.68 20.06 19.46 18.55 17.701.6 27.30 26.40 25.15 23.35 21.95 20.80 19.78 18.881.7 29.10 28.10 26.70 24.80 23.40 22.05 20.90 20.101.8 30.80 29.70 28.30 26.35 24.75 23.35 22.25 21.201.9 32.25 30.25 29.60 27.55 26.15 24.65 23.50 22.502.0 34.20 33.00 31.45 29.27 27.70 25.95 24.70 23.102.2 37.65 36.40 34.70 32.20 30.20 28.50 27.20 25.902.4 41.10 39.60 37.75 35.15 33.30 31.20 29.70 28.352.5 42.70 41.40 39.30 36.60 34.40 32.45 30.90 29.502.6 44.50 42.60 40.90 38.05 35.75 33.80 32.15 30.702.8 47.80 46.20 44.00 41.70 38.45 36.35 34.55 33.103.0 51.30 49.55 47.40 43.95 41.25 38.90 37.10 35.453.2 54.30 52.40 49.90 46.60 43.90 41.60 39.80 37.703.4 58.25 56.20 53.50 49.80 46.70 44.20 42.00 40.103.6 62.20 59.60 56.60 52.70 49.40 46.80 44.50 42.603.8 65.00 62.70 59.75 55.60 51.50 49.40 46.90 44.804.0 68.20 66.00 62.90 58.55 55.00 52.00 49.40 47.204.5 77.00 74.40 70.70 65.80 61.80 58.50 55.60 48.505.0 86.90 82.60 78.75 73.20 68.75 65.00 61.75 59.205.5 94.00 90.90 87.40 80.60 75.60 71.60 68.00 64.906.0 102.06 98.90 94.30 87.90 82.40 77.90 74.15 70.906.5 111.0 107.2 101.10 95.20 88.40 84.50 80.40 76.807.0 119.6 115.6 110.0 102.2 96.90 90.90 86.40 82.708.0 136.8 132.2 125.7 117.1 110.0 104.0 97.9 94.59.0 162.2 148.6 141.6 131.8 123.6 116.9 111.2 106.210.0 171.0 165.0 157.2 146.5 137.5 130.0 123.5 118.111.0 188.3 181.5 173.0 160.7 151.0 143.2 136.0 130.012.0 205.0 198.0 182.0 175.2 165.0 156.0 147.0 141.013.0 221.0 214.2 204.5 182.5 178.5 169.0 160.9 153.514.0 240.0 231.0 220.0 197.0 192.5 181.9 173.0 165.515.0 256.5 247.5 236.0 216.8 200.6 194.6 185.5 177.016.0 273.0 264.0 251.0 253.5 219.5 208.0 197.8 188.817.0 291.0 281.0 267.0 248.0 234.0 220.5 209.0 201.018.0 308.0 297.0 283.0 263.5 247.5 233.5 222.5 212.019.0 322.5 302.5 296.0 275.6 261.6 246.5 235.0 225.020.0 342.0 330.0 314.5 292.7 277.0 259.6 247.0 231.022.0 376.5 364.0 347.0 322.0 302.0 285.0 272.0 259.024.0 411.0 395.0 377.5 351.5 333.0 312.0 297.0 283.526.0 445.0 428.0 409.0 380.5 357.5 338.5 321.5 307.028.0 478.0 462.0 440.0 417.0 384.5 363.5 345.5 331.030.0 513.0 495.5 474.0 439.5 412.5 389.0 371.0 354.5

Diagram 4 shows the following relationship:

Determining velocity and nominal sizeVelocity w of the medium

Nominal size DN

AB 04 EN 11

Flow through pipelines

Diagram 4 · V.

– w diagram for gases, steam and liquids

0.1 0.2 0.3 0.6 10.1

0.2

0.30.40.50.60.8

1

4

2

3

10865

40

20

30

100806050

400

200

300

1000800600500

4000

2000

3000

10000

m3

h

800060005000

2 3 4 106 8 20 30 40 200 400

w

100600.4 0.8 30080

ms

20025

0

15012

5

100

8065

50

4032

2520

158

¼"

"

V.

pipe Volume flow rate in m3/hwpipe Flow velocity in m/sApipe Cross-sectional area of the assumed nominal

size plotted as a straight lineDN Nominal size

For gas service, the flow rate determined from Dia-gram 4 in m3/h can be converted into kg/h or m3/hfor the normal condition (refer to sizing sample prob-lem 14).

V.

G Gas flow rate, based on normal condition

V.

pipe = Apipe · wpipe (15)

wpipe = V.

pipe ·18.8DN

⎛⎝⎜

⎞⎠⎟

2(16)

DN = 18.8 ·V

w

.

pipe

(17)

wmax Self-operatedregulators1)

wmax District heating1) Nominal size DN in mm/inch

1) Only for water

Determine: Flow rates of compressed air for operating and normalconditions

Given: Pipe diameter · Pressure p in the pipe · Flow velocitySolution: Volume flow rate in m3/h for the operating condi-

tion can be obtained from Diagram 4 using DN 32 andwpipe · Determine ρ from Diagram 3 using p.Calculate the mass flow rate of the compressed air in kg/h.

V.

G is the volume flow rate of gas in m3/h for normal con-

dition at 0 °C and 1013 mbar. Determine V.

G accordingto the equation to the right.

Specifications subject to change without notice.

SAMSON AG · MESS- UND REGELTECHNIKWeismüllerstraße 3 · 60314 Frankfurt am Main · GermanyPhone: +49 69 4009-0 · Fax: +49 69 4009-1507Internet: http://www.samson.de AB 04 EN


W = V.

· ρ (18)

Nominal size = DN 32p = 5 barwpipe = 7 m/s

V.

= 20 m3/hFrom Diagram 3:

ρ = 6.3 kg/m3

W = 20 · 6.3 = 126 kg/h

Normal condition:1 m3 of air ⇒ 1.293 kg

V.

G = W1.293

= 1261293.

= 97.5 m3/h

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